Mathematical Literacy P1 Nov 2017 Memo Eng hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICAL LITERACY P1 NOVEMBER 2017 MARKING GUIDELINE MARKS: 150 SYMBOL EXPLANATION M Method MA Method with accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT/RG Reading from a table/graph/diagram SF Correct substitution in a formula O Opinion/Example/Definition/Explanation P Penalty, e.g. for no units/incorrect rounding off, etc. R Rounding off NPR No penalty rounding or omitting units AO Answer only, if correct, full marks This marking guideline consists of 16 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 2 DBE/November 2017 NSC – Marking Guideline Question 1 [30 MARKS] Ques Solution Explanation Topic/L F 1.1.1 D RT 2RT correct letter L1 (2) D 1.1.2 G  RT 2 RT correct letter L1 (2) M 1.1.3 C  RT 2 RT correct letter L1 (2) F 1.2.1 Profit = R18 700 – R 14 960 M 1M subtracting correct values L1 = R 3 740 A 1A calculating profit AO (2) M M 1.2.2 1M adding L1 10:15 + 5h50 = 16:05 A 1A correct time of sale 16:05 OR 4:05 pm OR 5 past 4 in the afternoon AO (2) M 1.2.3 Radius = 32,8 mm ÷ 2 MA 1MA dividing diameter by 2 L1 (a) = 16,4 mm CA 1CA radius AO (2) M 1.2.3 Distance = (71,8 mm – 32,8 mm) ÷ 2 MA 1MA subtracting and dividing L1 (b) = 19,5 mm CA 1CA distance OR OR 71,8 mm ÷ 2 = 35,9 mm MA 1MA subtracting and dividing Distance = 35,9 mm – 16,4 mm = 19,4 mm CA 1CA distance AO (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 3 DBE/November 2017 NSC – Marking Guideline Ques Solution Explanation Topic/L M 1.3.1 Cost of diluted juice per litre L1 = R 44,95 ÷ 14 ℓ MA 1MA dividing = R 3, 210714286 ≈ R 3,21 CA 1CA cost per litre NPR AO (2) M A 1.3.2 2 ℓ : 12 ℓ 1A correct volume of water and L1 order 1 : 6 CA 1CA simplification 1 Accept 6 AO (2) M 1.3.3 14 1MA dividing the correct L1 Number of glasses of juice = MA 0,175 values = 80 CA 1CA simplification to a whole number AO (2) RT MA D 1.4.1 35 39 39 60 63 84 93 107 117 120 126 142 1RT all values L1 1MA ascending order (2) D 1.4.2 July OR 7 month A th 2A correct month L1 (2) D 1.4.3 9 A 2A correct mode L1 (2) D th 1.4.4 April OR 4 month A 2A correct month L1 (2) A D 1.4.5 May and July A 1A May L1 OR 5th month and 7th month 1A July (2) [30] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 4 DBE/November 2017 NSC – Marking Guideline QUESTION 2 [46 MARKS] Ques Solution Explanation Topic/L F 2.1.1 R465,00 RT 2RT correct bus fare L1 (2) RT F RT 2.1.2 Queenstown and King William's Town 2RT correct cities L1 (2) F 2.1.3 Port Elizabeth to Bloemfontein = R435,00 RT 1RT R435 L1 (a) Cost = R755,00 – R435,00 = R320,00 CA 1CA cost Accept trial and error method AO (2) CA from Q2.1.3(a) F 2.1.3 King William's Town RT 2RT correct city L2 (b) (2) F 2.1.4 Cost excluding VAT L2 100M 1M × 100 = R365,00 × 114 M 1M ÷ 114 = R320,175… ≈ R320,18 CA 1CA simplification OR OR Cost excluding VAT R365 M 1M dividing = ≈ R320,18 CA 1,14 MA 1MA 1,14 1CA simplification OR OR 114 : 365 = 100 : x x = price excl. VAT M 1M proportion 100 x = R365,00 × M 1M x as subject of 114 formula = R320,175… ≈ R320,18 CA 1CA simplification OR OR 14 VAT = R365 × M = R44,82 1M multiplying with ratio 114 M CA 1M subtracting VAT Cost excluding VAT = R365 – R44,82 ≈ R320,18 1CA simplification NPR AO (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 5 DBE/November 2017 NSC – Marking Guideline Ques Solution Explanation Topic/L 2.1.5 From Queenstown to Bloemfontein return trip F RT 1RT correct fare L2 = R410 × 2 1CA for calculating the = R820 CA return trip Total travelling cost = 12 × R820 M 1M multiplying by 12 = R9 840 CA 1CA total cost OR OR Number of trips = 2 × 12 M 1M multiplying by 12 = 24 CA 1CA total trips Total travelling cost = 24 × R410 RT 1RT correct fare = R9 840 CA 1CA total cost OR OR One way cost for a year RT 1RT correct fare = R410 × 12 M 1M multiplying with 12 = R4 920 Total traveling cost 1M multiplying with 2 = R4 920 × 2 M 1CA total cost = R9 840 CA OR OR RT M 1RT correct fare Traveling cost = R410 × 2 × 12 M 1M multiplying with 2 1M multiplying with 12 =R9 840 CA 1CA cost AO (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 6 DBE/November 2017 NSC – Marking Guideline Ques Solution Explanation Topic/L RT 2.2.1 July 2013 OR 07/2013 OR 07/13 1RT month F RT 1RT year L1 (2) F 2.2.2 Water and Sewerage RT 1RT water and/or sewerage L1 1RT refuse Refuse Removal RT Penalty for including property rates (2) F 2.2.3 November = 3 days, December = 20 days M 1M adding L1 end date 2016/12/20 OR 20 December 2016 A 1A end date 20 Dec Accept 19 Dec AO (2) F 2.2.4 Daily average consumption L1 RT 1RT correct value = 12,00 kℓ ÷ 23 days M 1M dividing in correct order ≈ 0,522 kℓ OR OR Verifying the consumption rate per day: RT 1RT correct value = 12,00 kℓ ÷ 0,522 kℓ/day M 1M dividing in correct order ≈ 23 days OR OR 0,522 kℓ/day × 23 days M 1M multiplying ≈ 12,00kℓ A 1A volume (2) F 2.2.5 Water R 1R variable expense L1 The amount of water consumption is not the same every month. O 2O explanation clearly showing change (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 7 DBE/November 2017 NSC – Marking Guideline Ques Solution Explanation Topic/L RT 1RT all values from bill F 2.2.6 A = R690 000 × R0,0069160 ÷ 12 L1 (a) 1CA simplification = R397,67 CA Note value for B can be used to calculate A AO (2) F 2.2.6 B = R397,67 – R115,27 M 1M subtracting correct L1 (b) values = R282, 40 CA 1CA simplification OR OR B = R880,10 – R167,58 – R430,12 M 1M subtracting correct values = R282,40 CA 1CA simplification AO (2) F 2.2.7 R 298,36 RT 1RT correct values L1 Sewerage rate per m2 = 463 = R0,6444060475 A 1A simplification RT OR OR 2 463m : R 298,36 1RT Correct values 1m2 : R0, 6444… A 1A simplification NPR AO (2) F 2.2.8 R919,33 RT 2RT unpaid amount L1 (2) F 2.2.9 Rounding up A 2A Rounding up L1 OR OR A 1A rounding Rounding (off) to the nearest R10,00 1A nearest 10 rand OR OR 1A rounding A 1A nearest 100 rand Rounding (off) to the nearest R100,00 (2) F 2.3.1 Commission = 1,95% × £360,00 MA 1MA calculating % L1 = £7,02 A 1A commission in pound AO (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 8 DBE/November 2017 NSC – Marking Guideline Ques Solution Explanation Topic/L F 2.3.2 360 1MA conversion L2 £360,00 = M/A 0,05773 1A simplification = R6 235,9258.. A CA 1CA rounding ≈ R6 235,93 or R6 235 or R6 236 OR OR R1,00 £1 = 0,05773 = R17,32201628 MA 1MA conversion £360 = R17,32201628 × 360 = R62 35,925862 A 1A simplification ≈ R6 235,93 CA 1CA rounding OR OR R1,00 = £0,05773 x = £360,00 1A multiplying with 360 1 ×360 A 1MA conversion x =R 0,05773 MA 1CA rounding = R6 235,93CA NPR AO (3) 2.3.3 F Interest after 1 year = R5 000 × 6,3% 1M calculate interest for L2 = R315 M first year Amount after year 1 = R5 000 + R315 1A simplification = R5 315,00 A Interest for full 2nd year = R5 315 × 6,3% 1CA 2nd year amount ≈ R334,845 CA ∴ Interest for 12 year = R334,845 ÷ 2 1M half year interest = R167,42 M Value of the fixed deposit = R5 315 + R167,42 = R5 482,42 CA 1CA simplification OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 9 DBE/November 2017 NSC – Marking Guideline Ques Solution Explanation Topic/L Interest after 1 year = R5 000 × 6,3% 1M calculate interest = R315 M for first year Amount after year 1 = R5 000 + R315 1A simplification = R5 315,00 A 6,3% M 1M 2nd year rate Second year interest rate = 2 = 3,15% CA 1CA half year 1 interest Interest for 2 year = R5 315 × 3,15% ≈ R167,42 Value of the fixed deposit = R5 315 + R167,42 = R5 482,42 CA 1CA simplification OR OR 1M calculate amount Amount after year 1 = R5 000 (1 + 0,063) M for first year A 1A simplification = R5 315,00 CA 1CA 2nd year amount Value of fixed deposit after 1 years 1 2  0,063  = R5 3151 +  M 1M half year  2  1CA simplification ≈ R5 482,42 CA (5) [46] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 10 DBE/November 2017 NSC – Marking Guideline QUESTION 3 [21 MARKS] Ques Solution Explanation Topic/L M 3.1.1 Number of tables = 240 ÷ 8 = 30 A 1A correct number of tables L1 Number of balloons = 4 × 30 = 120 CA 1CA minimum number of balloons AO (2) M 3.1.2 Length of decorative ribbon in cm L2 = 2 × (length + width) + 1 2SF substituting correct values SF into the formula = 2 × (10 + 6) + 1 = 33 A 1A minimum length AO (3) M 3.1.3 Volume = π × (radius)2 × height L2 1A radius A SF = 3,142 × (6 cm)2 × 28 cm 1SF correct height and 3,142 = 3 167,136 cm3 CA 1CA simplification NPR (3) A M 3.1.4 Volume = 1 680 cm3 × 45% = 756 cm3 1A calculating 45% L2 Mass of sand = 756 cm3 × 1,53g/cm3 M 1M multiply by rate CA = 1 156,68 g ÷ 1 000 1CA mass in grams ≈ 1,16 kg C 1C converting to kg to 2decimal places OR OR 1,53 g/cm3 = 0,00153 kg/cm3 C 1C converting to kg A 1A calculating 45% Volume = 1 680 cm3 × 45% = 756 cm3 M 1M multiplying with the rate Mass of the sand = 0,00153 kg/cm3 × 756 cm3 = 1,15668 kg ≈ 1,16 kgCA 1 CA mass in kg to 2 dec. places OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 11 DBE/November 2017 NSC – Marking Guideline Ques Solution Explanation Topic/L Mass of sand in a full vase = 1 680 cm3 × 1,53g/cm3 M 1M multiplying with the = 2 570,4 g A rate 1A mass = 2,5704 kg C 1C conversion Mass of sand if filled to 45% = 2,5704 kg × 45% = 1,16 kg CA 1CA mass of sand to two decimal places (4) M A RT 3.2.1 1 1A substituting correct L2 Area of triangle = × 4 cm × 3,464 cm 2 values in formula 1RT height = 6,928 cm2 CA 1CA simplification NPR AO (3) CA from Q3.2.1 M 3.2.2 Total surface Area of a triangular prism L3 CA SF 1CA substituting area of = 2 × 6,928 + 3 × 6 cm × 4cm triangle 1SF substituting correct = 13,856 cm2 + 72 cm2 CA values in formula 1CA simplification = 85,856 cm2 CA 1CA total surface area (4) M 3.2.3 30 minutes = 1 800 seconds C 1 C conversion to seconds L1 1 800 Average time to cover 1 box = seconds 20 = 90 seconds CA 1CA simplification OR Average time to cover 1 box OR 30 min = = 1,5 min M 20 1M time per box = 1,5 min × 60 sec/min = 90 seconds C 1C conversion AO (2) [21] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 12 DBE/November 2017 NSC – Marking Guideline QUESTION 4 [27 MARKS] NOTE :MPU & NC maximum [23 MARKS] to be scaled to 27 MARKS Ques Solution Explanation Topic/L A M&P 4.1.1 Bar scale OR Scaled bar OR Linear scale OR 2A identifying type of L1 Graphical scale scale (2) M&P 4.1.2 Top view OR Aerial view OR Bird’s eye view A 2A correct view of the L1 OR Satelite view map (2) A M&P 4.1.3 South East OR SE OR East of South 2A identifying correct L1 direction (2) M&P 4.1.4 5 A 2A exact number of L2 medical points Accept 4 (2) A M&P A 4.1.5 Mowbray and Observatory 2A identifying correct L1 suburbs Accept Maitland and Saltriver (2) A A A M&P 4.1.6 Castle De Goede Hoop, Old Biscuit Mill , Planetarium 3A identifying correct L2 OR 4, 5 and 6 tourist attractions (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 13 DBE/November 2017 NSC – Marking Guideline Ques Solution Explanation Topic/L NOTE: M&P 4.2.1 D; B; E; A; C A [MPU & NC not to be L2 marked] A 1A order BEA 1A end with C (2) NOTE: M&P 4.2.2 E OR B A [MPU & NC not to be L1 marked] 2A correct letter (2) A P 4.2.3 0 2A probability L2 0 % OR Impossible OR 0 OR OR None (a) 130 (2) P 4.2.3 Total blocks = 20 + 25 + 28 + 30 + 27 = 130 A 1A total 130 L2 (b) Probability of taking out a blue block 25 A 1A numerator = 130 A 1A denominator 5 OR OR 19,23% OR 0,19 26 AO (3) MA M&P 4.2.4 Number of layers = 35 cm ÷ 16, 1MA dividing correct values L1 (a) = 2,12… ≈ 2 CA 1CA exact number of layers AO (2) M&P 4.2.4 Number of cans which can be packed lengthwise L3 (b) = 56 cm ÷ 12,6 cm MA 1MA dividing the width or length by 2,6 = 4,444… ≈ 4 Number of cans which can be packed width-wise = 41 cm ÷ 12,6 cm 1A rounding both down to = 3,253… ≈ 3 A whole numbers Maximum number of cans = 4×3×2 = 24 CA 1CA for max number of cans AO (3) [27] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 14 DBE/November 2017 NSC – Marking Guideline QUESTION 5 [26 MARKS] Ques Solution Explanation T/L D 5.1.1 Broken line graph OR line graph A 2A correct type of graph L1 (2) M D 5.1.2 Number of candidates = 287 453 + 389 615 1M adding Math and Math Lit L2 = 677 068 CA 1CA max number of candidates AO (2) P 5.1.3 100% OR 1 OR certain OR definite A 2A correct probability L2 (2) st RT RT 1RT 1 subject D RT 5.1.4 Accounting, Business Studies, Economics and 1RT 2nd subject L1 Mathematical Literacy 1RT last two subjects (3) D 5.1.5 Mathematics RT 2RT correct subject L1 (2) A D 5.1.6 The data of one variable is grouped into subjects 2A explanation L1 OR (2) The data of one variable is not numerical A D 5.1.7 Business Studies RT 2RT correct subject L1 (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 15 DBE/November 2017 NSC – Marking Guideline Ques Solution Explanation T/L D O 5.2.1 Copyright payments, advertising costs, bursary, 2O an example of other type of L1 grants etc. expenditure (OR any other valid expenditure) (2) D 5.2.2 Donations M L2 = [R63 – (R27,09 + R21,02 + R3,78)] billion 1M subtracting from R63 billion = R11,11 billion CA 1CA simplification in billions 11,11 Percentage donations = × 100% 63 ≈ 17,6% CA 1CA donations as a % OR OR R27,09 + 21,02 + 3,78 = R51,89 billion Percentage income shown R51,89 = × 100% R63 ≈ 82,4% M 1M percentage income shown Percentage donations 1M subtracting from 100% =100% – 82,4% M =17,6% CA 1CA simplification OR OR Percentage R27,09 = × 100% = 43% M 1M percentage calculation R63 R21,02 × 100% ≈ 33,365% R63 R3,78 × 100% = 6% R63 Percentage donations = 100% – (43% + 33,4% + 6%) M 1M subtracting from 100% = 17,6% CA 1CA simplification NPR AO (3) Copyright reserved

Downloaded from hlayiso.com Mathematical Literacy/P1 16 DBE/November 2017 NSC – Marking Guideline Ques Solution Explanation T/L RT F 5.2.3 Interest in Rand = 54 100 000 000 × 0,7% M 1RT correct amount L1 CA 1M multiplying with 0,7% = 378 700 000 OR 378,7 million 1CA interest amount OR OR RT Interest in rand = 54,1 billion × 0,7% M 1RT correct amount 1M multiplying with 0,7% = 0,3787 billion CA 1CA interest amount = 378 700 000 OR 378,7 million AO (3) D 5.2.4 Difference = income – expenditure L2 M 1M subtracting = R63 billion – R54,1 billion = R8,9 billion CA 1CA simplification in billions C = R8 900 million OR R8 900 000 000 1C for difference in millions OR OR Difference = income – expenditure 1M subtracting M C 1C converting to millions = R63 000 million – R54 100 million CA = R8 900 million OR R8 900 000 000 1CA difference in millions (3) [26] TOTAL: 150 Copyright reserved